An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 1.5 x 10-7 C/m2, and the plates are separated by a distance of 1.4 x 10-2 m. How fast is the electron moving just before it reaches the positive plate?
THANK YOU for any help or hints or anything that you can give me!
I would find the voltage on the capacitor, or Electric field first. Using Gauss Law, then E= sigma/2epsilion
Force/q= E
Force=Eq
force*distance=Eq*distance
1/2 m v^2=Eq*distance
solve for velocity v.
The E field between the plates is proportional to the charge per unit area, sigma (1.5 x 10-7 C/m2).
E = sigma/epsilon
where epsilon is the permittivity of free space, which you should look up.
The kinetic energy that is given to the electrion while crossing the gap of length X is
KE = e E X
Set that equal to the final kinetic energy and solve for the velocity.
Dr WLS is correct, E between the plates is sigma/episilon, not sigma/2epsilon.
To find the speed of the electron just before it reaches the positive plate, we can use the concept of conservation of energy. The electric potential energy gained by the electron as it moves towards the positive plate is converted into kinetic energy. We can equate the electric potential energy with the kinetic energy to solve for the speed.
1. First, let's find the electric potential energy gained by the electron. The electric potential energy is given by the equation:
PE = qV
where PE is the potential energy, q is the charge, and V is the voltage. In this case, since the electron is moving towards the positive plate, the voltage is the potential difference between the plates.
V = Ed
where E is the electric field strength and d is the distance between the plates.
E = σ/ε
where σ is the charge per unit area on each plate and ε is the permittivity of free space.
Substituting these values, we get:
V = (σ/ε) * d
PE = q * (σ/ε) * d
2. Next, let's find the kinetic energy of the electron just before it reaches the positive plate. The kinetic energy is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the electron, and v is its speed.
3. Since we are assuming the electron starts from rest, the initial kinetic energy is zero. Therefore, we can equate the electric potential energy gained to the final kinetic energy:
PE = KE
q * (σ/ε) * d = (1/2) * m * v^2
4. Now, we can solve for the speed of the electron by rearranging the equation and plugging in the known values:
v = sqrt((2 * q * (σ/ε) * d) / m)
Plug in the values:
q = charge of an electron = -1.6 x 10^-19 C
σ = charge per unit area on each plate = 1.5 x 10^-7 C/m²
ε = permittivity of free space = 8.85 x 10^-12 C²/(Nm²)
d = distance between the plates = 1.4 x 10^-2 m
m = mass of an electron = 9.11 x 10^-31 kg
Calculate the speed using the equation:
v = sqrt((2 * (-1.6 x 10^-19) * (1.5 x 10^-7) * (1.4 x 10^-2)) / (9.11 x 10^-31))
The final answer will give you the speed of the electron just before it reaches the positive plate.